As the diode ${\mathrm{D}}_2$ is connected in reverse order, there is no current through it.

The effective resistances of the circuit can, then, be calculated as follows:

$\begin{array}{rcl}{R}_{eff}& =& 25+\frac{{\left(25+125\right)}^2}{\left(25+125\right)+\left(25+125\right)} \mathrm{\Omega }\\ & =& 100 \mathrm{\Omega }\end{array}$

Each current through the diodes can be calculated as follows:

$\begin{array}{rcl}{I}_1& =& \left(\frac{5 \mathrm{V}}{100 \mathrm{\Omega }}\right)\\ & =& 0.05 \mathrm{A}\end{array}$

As the diode ${\mathrm{D}}_2$ is non-conducting, the current $I_1$ will be divided equally among the other two diodes ${\mathrm{D}}_1$ and ${\mathrm{D}}_3$.

Hence, the currents in the diodes can be written as

$\begin{array}{rcl}{I}_2& =& \frac{0.05 \mathrm{A}}{2}\\ & =& 0.025 \mathrm{A}\\ I_4& =& 0.025 \mathrm{A}\end{array}$

Thus,

$\begin{array}{rcl}\frac{I_1}{I_2}& =& \frac{0.05 \mathrm{A}}{0.025 \mathrm{A}}\\ & =& 2\end{array}$